Locus of the centre of the circle which touches the two circle $x^{2} + y^{2} + 8 x - 9 = 0 a n d x^{2} + y^{2} - 8 x + 7 = 0$ externally, is

$x^{2} + y^{2} = 0$

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$\frac{x^{2}}{1} + \frac{y^{2}}{15} = 1$

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$15 x^{2} + 2 x - 1 = 0$

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$\frac{x^{2}}{1} - \frac{y^{2}}{15} = 1$

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Solution

The correct option is D
$\frac{x^{2}}{1} - \frac{y^{2}}{15} = 1$

As, $| c c_{1} - c c_{2} | = | (r + r_{1}) - (r + r_{2}) | = c o n s t a n t w h e r e | r_{1} - r_{2} | < c_{1} c_{2} \Rightarrow L o c u s o f C i s a h y p e r b o l a w i t h f o c i c_{1} a n d c_{2} i . e, (- 4, 0) a n d (4, 0) a l s o, 2 a = | r_{1} - r_{2} | = 1 \Rightarrow a = 1 N o w, e = \frac{2 a e}{2 a} = \frac{8}{2} = 4 s o, b^{2} = 1^{2} (4^{2} - 1 = 15$
Hence, Locus of centres of circle is hyperbola, whose equation is
$\frac{x^{2}}{1} - \frac{y^{2}}{15} = 1$

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Locus of the centre of the circle which touches the two circle x2+y2+8x–9=0 and x2+y2–8x+7=0 externally, is

Locus of the centre of the circle which touches the two circle $x^{2} + y^{2} + 8 x - 9 = 0 a n d x^{2} + y^{2} - 8 x + 7 = 0$ externally, is